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AysviL [449]
3 years ago
14

Item 12 You win an online auction for a toy. Your winning bid of $52 is 65% of your maximum bid. How much more were you willing

to pay for the toy than you actually paid?
Mathematics
1 answer:
nataly862011 [7]3 years ago
5 0

Answer:

$28

Step-by-step explanation:

Given that:

Value of Winning bid = $52

Winning bid 65% of the maximum bid.

To find:

How much more is the maximum bid from the winning bid ?

Solution:

We are given that the winning bid is 65% of the maximum bid.

Using this percentage value, we need to first find the value of maximum bid and then we need to subtract the value of winning bid from the maximum bid to find the answer.

Let the value of maximum bid = $x

As per question statement:

65%\ of\ x = 52\\\Rightarrow \dfrac{65}{100} \times x =52\\\Rightarrow 5x = 400\\\Rightarrow x = \dfrac{400}{5}\\\Rightarrow x =\$80

Therefore, maximum bid = $80

Our answer is:

$80 - $52 = <em>$28</em>

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Answer:

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6 0
3 years ago
What is the answer to this? PLEASE HELP!
Leno4ka [110]
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What is 25.691 rounded to the greatest place
kvv77 [185]
The greatest place value is the tens spot which 2 is in the tens spot so we need to round up since 5 is next to the 2. The answer would be 30
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3 years ago
The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi
eimsori [14]

Answer:

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts  and data given  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X=4.2 represent the sample mean  

s=1.4 represent the sample standard deviation  

n=16 represent the sample selected  

\alpha=0.05 significance level  

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:    

Null hypothesis:\mu = 3.6    

Alternative hypothesis:\mu \neq 3.6    

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic  

We can replace in formula (1) the info given like this:    

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with df=n-1=16-1=15 degrees of freedom. The value for \alpha=0.05 and \alpha/2=0.025 so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got t_{crit}=\pm 2.131    

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

Conclusion    

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

3 0
3 years ago
Help ASAP!! Geometry -
horrorfan [7]

Answer:

x = 9.27 cm

Volume = 289.2 cm³

Step-by-step explanation:

SA = Surface Area

SA = 229.2 cm²

P = Perimeter of Base

P = 3(6)

P = 18 cm

H = Height of Solid

H = x

A = Area of Base(Equilateral Triangle)

A = 6²√3 / 4

A = 36√3 / 4

A = 9√3

A = 9(1.732)

A = 31.176

A = 31.2 cm²

SA = Surface Area

SA = PH + 2A

229.2 = 18x + 2(31.2)

229.2 = 18x + 62.4

229.2 - 62.4 = 18x

166.8 = 18x

9.266 = x

9.27 = x

x = 9.27 cm

V = Volume

V = AH

V = 31.2(9.27)

V = 289.224

V = 289.2 cm³

8 0
3 years ago
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